Question

A ball is projected in horizontal direction from top
of inclined plane of inclination α. Time after which
it strikes the inclined plane is T. The maximum
distance of the ball from the inclined surface is​

Answer 1

maximum distance of ball from the inclined surface is gT²/2tanθ.

ball is projected in horizontal direction from the top of inclined plane.

see figure,

let ball strikes the plane h height below from the top of inclined plane where distance of striking point from the inclined plane is x.

so, h/x = tanθ

h = xtanθ.....(1)

now apply formula, $y=u_yt+\frac{1}{2}a_yt^2$

here, y = -h, uy = 0, ay = -g and t = T

so, -h = 0 - 1/2 gT²

or, xtanθ = 1/2 gT²

or, x = gT²/2tanθ

hence maximum distance of ball from the inclined surface is gT²/2tanθ.

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Answer 2

$\huge\bold\purple{Answer:-}$

ball is projected in horizontal direction from the top of inclined plane.

see figure,

let ball strikes the plane h height below from the top of inclined plane where distance of striking point from the inclined plane is x.

so, h/x = tanθ

h = xtanθ.....(1)

now apply formula, y=u_yt+\frac{1}{2}a_yt^2

here, y = -h, uy = 0, ay = -g and t = T

so, -h = 0 - 1/2 gT²

or, xtanθ = 1/2 gT²

or, x = gT²/2tanθ

hence maximum distance of ball from the inclined surface is gT²/2tanθ.